Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 4x}{x + 6} = \dfrac{-5x - 18}{x + 6}$
Answer: Multiply both sides by $x + 6$ $ \dfrac{x^2 + 4x}{x + 6} (x + 6) = \dfrac{-5x - 18}{x + 6} (x + 6)$ $ x^2 + 4x = -5x - 18$ Subtract $-5x - 18$ from both sides: $ x^2 + 4x - (-5x - 18) = -5x - 18 - (-5x - 18)$ $ x^2 + 4x + 5x + 18 = 0$ $ x^2 + 9x + 18 = 0$ Factor the expression: $ (x + 6)(x + 3) = 0$ Therefore $x = -6$ or $x = -3$ At $x = -6$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -6$, it is an extraneous solution.